# Wednesday November 20

Last time:
$$
\ZZ \Lambda \iff \theset{ \lieh^* \to \ZZ_{\geq 0} \mid \sim  } \\
e(\mu) \mapsto e_\mu \\
e(\lambda) e(\mu) = e(\lambda + \mu) \mapsto f \star g(\lambda) = \sum_{a+b=\lambda} f(a) g(b)
$$

and $\mathrm{ch} L(\lambda) = \sum_{\mu \in \Lambda} \dim L(\lambda)_\mu e(\mu)$.

We have the Kostant function $p(\lambda) = \# \theset{ (k_\alpha)_\alpha \mid -\lambda = \sum_{\alpha\in\Phi^+} k_\alpha \alpha  }$ 
and the Weyl function $q = e_\rho \star \prod_{\alpha\in\Phi^+}(1 - e_{-\alpha}) = \prod_{\alpha\in\Phi^+} (e_{\alpha/2} - e_{-\alpha/2})$.

**Lemma:**
$p\star e_\lambda = \mathrm{ch} M(\lambda)$, so $q \star \mathrm{ch} M(\lambda) = e_{\lambda + \rho}$ and $q \star p = e_\rho$.

## Weyl's Character Formula  (24.2-3)

**Definition:** 
The *dot action* of $W$ is given by $w\cdot \lambda = w(\lambda + \rho) - \rho$, i.e. a reflection for hyperplanes passing through $-\rho$.

E.g. for type $A2$, where $W(0) = 0$, we have:

![Type A2](figures/2019-11-20-09:19.png)\

And for the dot action, we have

![Image](figures/2019-11-20-09:22.png)\

where $W \cdot 0 = 0$ and $s(\alpha_1) = -\alpha_1$.

**Theorem (Harish-Chandra):**
If $L(\mu)$ is a composition factor of $M(\lambda)$, then $\mu \in W\cdot \lambda$ for $\mu \leq \lambda$.

*Proof:*
Postponed.

> $\mathrm{ch}$ are characters, $L(\lambda)$ is a Verma module.

*Remark:* 
If we sum over $\mu \leq \lambda$, we obtain

\begin{align*}
\mathrm{ch} M(\lambda) &= \sum_{\mu \in W \cdot \lambda} a_{\lambda \mu} \mathrm{ch} L(\mu) \\
\mathrm{ch} L(\lambda) &= \sum_{\mu \in W \cdot \lambda} b_{\lambda \mu} \mathrm{ch} M(\mu) \\
&= \sum_{W\cdot \lambda \in \Lambda} c_{\lambda W} \mathrm{ch} M(w\cdot \lambda)
.\end{align*}

**Theorem (Weyl's Character Formula):**
If $\lambda \in \Lambda^+$, then 
$$
\mathrm{ch} L(\lambda) = \frac{ \sum_{w\in W} (-1)^{\ell(w)} e(w \cdot \lambda) }{ \sum_{w\in W} (-1)^{\ell(w)} e(w\cdot 0) }
$$

*Proof*:

We have $\mathrm{ch} L(\lambda) = \sum_{w} c_{\lambda w} \mathrm{ch} M(w\cdot \lambda)$, and so by the lemma, 
$$
q \ast \mathrm{ch} L(\lambda) = \sum c_{\lambda w} q \star \mathrm{ch} M(W(\lambda + \rho) - \rho) = \sum_w c_{\lambda w} e_{W(\lambda + p)}
$$

Thus for all $\alpha \in \Phi^+$, we have
$$
s_\alpha(q \star \mathrm{ch} L(\lambda)) = \sum_w c_{\lambda, s_\alpha w} e_{w(\lambda + \rho)}
$$

On the other hand, by part (c) of the lemma, we have 
$$
(s_\alpha \star q) \star \mathrm{ch} L(\lambda) 
= -q \star \mathrm{ch} L(\lambda) 
= \sum_w -c_{\lambda, w} e_{w(\lambda+\rho)}
$$

which implies that $c_{\lambda, s_\alpha w} = -c_{\lambda, w}$ by comparing term-by-term, and thus $c_{\lambda, w} = (-1)^{\ell(w)}$ because $c_{\lambda e} = 1$.

In particular, $q = q \star e(0) = q \star \mathrm{ch} L(0) = \sum_{w\in W} (-1)^{\ell(w)} e_{w(\rho)}$, and thus

\begin{align*}
\mathrm{ch} L(\lambda) = 
\frac {
\sum_w (-1)^{\ell(w)} e_{w(\lambda+p)}
}{
\sum_w (-1)^{\ell(w)} e_{w(p)}
} \\
=
\frac {
\sum_w (-1)^{\ell(w)} e(w\cdot \lambda) 
}{
\sum_w (-1)^{\ell(w)} e(w \cdot 0)
}.
\end{align*}


$\qed$

*Example:* 
For type $A1$, we have $W = \Sigma_2 = \theset{\mathbf{1}, s}$.
Take $\lambda = 3$ under

\begin{align*}
\Lambda \equiv \ZZ \\
\alpha_1 \to 2 \\
w_1 = \rho \to 1
,\end{align*}


from which we obtain 

\begin{align*}
\mathrm{ch} L(3) = 
\frac {
e(\mathbf{1} \cdot 3) - e(s\cdot 3)
}{
e(\mathbf{1} \cdot 0) - e(s\cdot 0)
} \\
=
\frac{e(3) - e(-5)}{e(0) - e(-2)} \\
= e(3) + e(1) + e(-1) + e(-3) \quad\text{by long division}
.\end{align*}

**Corollary (Kostant's Dimension Formula):**

If $\mu \leq \lambda \in \Lambda^+$, then 
$$
\dim L(\lambda)_\mu = \sum_{w\in W} (-1)^{\ell(w)} P(w\cdot \lambda - \mu).
$$

*Proof:*
$p\star e_\mu(w \cdot \lambda) = \sum_{a+b = w\cdot \lambda} p(a) e_\mu(h) = p(w\cdot \lambda - \mu)$, since this is the only term that survives.

Then $p(w\cdot \lambda - \mu)$ is the coefficient for $e(\mu)$ in $\mathrm{ch} M(w\cdot \lambda) = \dim M(\lambda)_\mu$.
Thus $\dim L(\lambda)_\mu = \sum_{w\in W} (-1)^{\ell(w)} \dim M(w\cdot \lambda)_\mu$.


**Corollary (Weyl's Dimension Formula):**

If $\lambda \in \Lambda^+$, then 
$$
\dim L(\lambda) = 
\frac{
\prod_{\alpha\in\Phi^+} (\lambda+\rho, \alpha\dual)
}{
\prod_{\alpha\in\Phi^+} (\rho, \alpha\dual)
}
$$

*Proof (sketch)*:

Define an operator $\del = \prod_{\alpha\in\Phi^+} \del_a$, where $\del_a: e(\mu) \mapsto (u, \alpha\dual) e(\mu)$.
Then $\del$ is well-defined since $\del_\alpha \del_\beta = \del_\beta \del_\alpha$ for all $\alpha, \beta$, and (exercise) $\del$ is a derivation.

Define an evaluation homomorphism $\nu: \sum_\mu c_\mu e(\mu) \mapsto \prod_\mu c_\mu$. 
Note that $\nu (\mathrm{ch} L(\lambda)) = \dim L(\lambda)$, and $\nu(q) = 0$ because $\nu(e_{\alpha_i - 1}) = 0$.

*Claim:*
$$
\nu(
\del(
q \star \mathrm{ch} L(\mu - \rho)
)
) = 
\abs{w} \prod_{\alpha\in\Phi^+} (\mu, \alpha\dual)
$$

This is relatively straightforward once you know that you have a derivation and a homomorphism.

With this claim, we have
$$
\nu(\del(q \star \mathrm{ch} L(\lambda))) = \nu(\del q) \nu(\mathrm{ch} L(\lambda)) + \nu(q) \nu(\del \mathrm{ch} L(\lambda))
$$

where we can identify a number of terms, and then taking ratios yields Weyl's dimension formula.